0
$\begingroup$

Regarding sequences of functions $(f_n(x))$, I can wrap my head around the idea that uniform convergence $\Rightarrow$ pointwise convergence, but pointwise convergence does not imply uniform convergence.

However, regarding series of functions $\sum f_n(x)$, I am not sure if I totally understand. I have proved that if a series converges uniformly on $\mathbb R$, then it is pointwise convergence on $\mathbb R$ and $\sup |f_n(x)|$ converges to 0 as n goes to infinity. But I am not sure of the converse. I was pretty sure the converse did not hold true, but I cannot find a counter example.

Is it similar to sequences and the converse does not hold true? Or am I thinking of it wrong and there is a way to prove that it does also hold true?

$\endgroup$
8
  • 2
    $\begingroup$ $\sum ((-1)^{n}/n)|x|$ converges pointwise on $\mathbb R$ but not uniformly. $\endgroup$ Commented May 17, 2021 at 8:46
  • $\begingroup$ How did I not think of that! It's a great counterexample. Thanks! $\endgroup$ Commented May 17, 2021 at 8:54
  • $\begingroup$ It is not true that uniform convergence $\Rightarrow \limsup f_n(x) = 0$, or that $\limsup f_n(x) = 0\Rightarrow$ uniform convergence. $\endgroup$
    – TonyK
    Commented May 17, 2021 at 9:04
  • 1
    $\begingroup$ The sequence $(99,99,99,\ldots)$ converges uniformly but not to zero. (Having said that, it's not clear to me what $\limsup f_n(x) = 0$ even means; perhaps you could clarify that?) $\endgroup$
    – TonyK
    Commented May 17, 2021 at 9:21
  • 1
    $\begingroup$ There is really no difference between sequences and series, because you can convert between them by simply defining the sequence $(g_n(x))$ as $g_n(x)=\sum_{i=0}^nf_n(x)$. Then all those theorems about uniform convergence become equivalent. $\endgroup$
    – TonyK
    Commented May 17, 2021 at 20:29

1 Answer 1

2
$\begingroup$

The series $\sum_{n=0}^\infty x^n$ converges pointwise to $\frac1{1-x}$ on $(-1,1)$. However, the convergence is not uniform, since each function $\sum_{n=0}^N x^n$ is bounded on $(-1,1)$, but $x\mapsto\frac1{1-x}$ isn't.

$\endgroup$
2
  • $\begingroup$ Ah. That makes sense. So a counter example exists since $\sum x^n$ converges uniformly only on a bounded interval. Thank you! $\endgroup$ Commented May 17, 2021 at 8:52
  • $\begingroup$ @interestingmeatloaf: That's not right $-$ $(0,1)$ is bounded. (I suppose that means that strictly speaking, what you wrote is correct. For instance, $\sum x^n$ doesn't converge on the unbounded interval $(99,\infty)$. But it's not what you meant!) $\endgroup$
    – TonyK
    Commented May 17, 2021 at 9:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .